Question: You have found the following ages (in years) of all 4 gorillas at your local zoo: $ 10,\enspace 4,\enspace 5,\enspace 6$ What is the average age of the gorillas at your zoo? What is the standard deviation? You may round your answers to the nearest tenth.
Solution: Because we have data for all 4 gorillas at the zoo, we are able to calculate the population mean $({\mu})$ and population standard deviation $({\sigma})$ To find the population mean , add up the values of all $4$ ages and divide by $4$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{4}} x_i}{{4}} $ $ {\mu} = \dfrac{10 + 4 + 5 + 6}{{4}} = {6.3\text{ years old}} $ Find the squared deviations from the mean for each gorilla. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $10$ years $3.7$ years $13.69$ years $^2$ $4$ years $-2.3$ years $5.29$ years $^2$ $5$ years $-1.3$ years $1.69$ years $^2$ $6$ years $-0.3$ years $0.09$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{13.69} + {5.29} + {1.69} + {0.09}} {{4}} $ $ {\sigma^2} = \dfrac{{20.76}}{{4}} = {5.19\text{ years}^2} $ As you might guess from the notation, the population standard deviation $({\sigma})$ is found by taking the square root of the population variance $({\sigma^2})$ ${\sigma} = \sqrt{{\sigma^2}}$ $ {\sigma} = \sqrt{{5.19\text{ years}^2}} = {2.3\text{ years}} $ The average gorilla at the zoo is 6.3 years old. There is a standard deviation of 2.3 years.